Question

you have a 1.53×10³ ppm solution. what is the concentration in molarity? solute is nacl. (density of the solution is 1.01 g/ml)

Explanation:

Step1: Understand ppm and Molarity

ppm (parts per million) for a solution can be considered as mg of solute per kg of solution (since 1 ppm = 1 mg/kg for dilute solutions, and here we can also use mg/L approximately, but with density, we'll be precise). Molarity (M) is moles of solute per liter of solution. First, let's assume we have 1 kg (1000 g) of solution. But with density, we can find volume. Alternatively, let's take a volume of solution, say 1 L (1000 mL) of solution.

Step2: Calculate mass of solution

Density ($
ho$) = 1.01 g/mL, volume (V) = 1000 mL. Mass of solution ($m_{solution}$) = $
ho \times V$ = $1.01 \, g/mL \times 1000 \, mL$ = $1010 \, g$ = $1.01 \, kg$.

Step3: Calculate mass of solute from ppm

ppm = $\frac{mass_{solute} (mg)}{mass_{solution} (kg)}$, so $mass_{solute} (mg)$ = ppm $\times mass_{solution} (kg)$. Given ppm = $1.53 \times 10^3$ ppm, $mass_{solution} = 1.01 \, kg$. So $mass_{solute} (mg)$ = $1.53 \times 10^3 \, mg/kg \times 1.01 \, kg$ = $1.53 \times 1.01 \times 10^3 \, mg$ = $1545.3 \, mg$ = $1.5453 \, g$.

Step4: Calculate moles of NaCl

Molar mass of NaCl: Na (22.99 g/mol) + Cl (35.45 g/mol) = 58.44 g/mol. Moles of NaCl ($n$) = $\frac{mass_{solute}}{molar \, mass}$ = $\frac{1.5453 \, g}{58.44 \, g/mol}$ ≈ $0.02644 \, mol$. Wait, no, wait. Wait, if we take 1 L of solution, mass of solution is 1010 g. ppm is also mg/L (since 1 ppm = 1 mg/L for water, but with density, let's re-express. Wait, ppm is mass of solute per mass of solution times $10^6$. So $ppm = \frac{mass_{solute}}{mass_{solution}} \times 10^6$. So $mass_{solute} = \frac{ppm \times mass_{solution}}{10^6}$. Let's take $mass_{solution} = 1010 \, g$ (1 L of solution). Then $mass_{solute} = \frac{1.53 \times 10^3 \times 1010 \, g}{10^6}$ = $\frac{1.53 \times 10^3 \times 1.01 \times 10^3 \, g}{10^6}$ = $\frac{1.53 \times 1.01 \times 10^6 \, g}{10^6}$ = $1.53 \times 1.01 \, g$ ≈ $1.5453 \, g$ (same as before).

Step5: Calculate moles of NaCl

Molar mass of NaCl ($M_{NaCl}$) = 58.44 g/mol. Moles of NaCl ($n$) = $\frac{mass_{solute}}{M_{NaCl}}$ = $\frac{1.5453 \, g}{58.44 \, g/mol}$ ≈ $0.02644 \, mol$? Wait, no, wait. Wait, if we take 1 L of solution, mass of solution is 1010 g. ppm is 1.53e3 ppm, which is 1.53e3 mg/kg = 1.53e3 mg/1000 g = 1.53 mg/g. So in 1010 g of solution, mass of solute is 1.53 mg/g * 1010 g = 1545.3 mg = 1.5453 g. Then moles of NaCl = 1.5453 g / 58.44 g/mol ≈ 0.02644 mol? Wait, no, that can't be right. Wait, no, ppm is parts per million by mass, so 1 ppm = 1 g solute per 1e6 g solution. So let's correct:

ppm = (mass solute / mass solution) * 1e6. So mass solute = (ppm * mass solution) / 1e6. Let's take mass solution = 1 L of solution, mass solution = 1010 g. So mass solute = (1.53e3 * 1010 g) / 1e6 = (1.53e3 * 1.01e3 g) / 1e6 = (1.53 * 1.01 * 1e6 g) / 1e6 = 1.53 * 1.01 g = 1.5453 g. Wait, that's the same as before. But wait, 1.53e3 ppm is 1530 ppm, which is 1530 mg/kg, so in 1 kg (1000 g) of solution, solute is 1530 mg = 1.53 g. But with density 1.01 g/mL, 1 kg of solution has volume = 1000 g / 1.01 g/mL ≈ 990.099 mL ≈ 0.9901 L. Then molarity would be moles / volume. Let's do it properly:

Let’s take V = 1 L of solution.

Mass of solution: 1.01 g/mL * 1000 mL = 1010 g.

Mass of solute: (1.53e3 ppm) * (1010 g solution) / 1e6 = (1.53e3 * 1010) / 1e6 g = (1.53 * 1.01) g = 1.5453 g (since 1e3 * 1010 = 1.01e6, so 1.53e3 * 1010 = 1.53 * 1.01e6, divided by 1e6 is 1.53*1.01).

Moles of NaCl: 1.5453 g / 58.44 g/mol ≈ 0.02644 mol? Wait, no, that's too low. Wait, no, I think I messed up the ppm interpretation. ppm for solutions can also be mg/L when density is 1 g/mL, but when density is different, we need to be careful. Wait, 1 ppm = 1 mg/L only when density is 1 g/mL (since 1 L = 1 kg). But when density is $
ho$ g/mL, 1 L of solution has mass $
ho$ kg. So ppm = (mg solute) / (kg solution) = (mg solute) / ($
ho$ kg) (since 1 L solution is $
ho$ kg). So mg solute = ppm * $
ho$ kg. So for 1 L of solution, mass of solution is $
ho$ kg (since 1 L = 1000 mL, mass = $
ho$ * 1000 g = $
ho$ kg). So mg solute = ppm * $
ho$ kg. Then g solute = (ppm * $
ho$) / 1000 g (since 1 mg = 0.001 g). Then moles of solute = (ppm * $
ho$) / (1000 * Molar mass). Then molarity = moles / 1 L = (ppm * $
ho$) / (1000 * Molar mass).

Let's use this formula. ppm = 1.53e3, $
ho$ = 1.01 g/mL = 1.01 kg/L, Molar mass of NaCl (M) = 58.44 g/mol.

So Molarity (M) = $\frac{ppm \times
ho}{1000 \times M}$

Plugging in: ppm = 1530, $
ho$ = 1.01 kg/L, M = 58.44 g/mol.

M = $\frac{1530 \times 1.01}{1000 \times 58.44}$

Calculate numerator: 1530 * 1.01 = 1545.3

Denominator: 1000 * 58.44 = 58440

M = 1545.3 / 58440 ≈ 0.02644? Wait, that can't be right. Wait, no, I think the mistake is in the ppm interpretation. Wait, 1 ppm is 1 g solute per 1e6 g solution. So if we have 1e6 g (1000 kg) of solution, solute is 1 g. But we have 1 L of solution, mass 1010 g. So mass solute = (1.53e3 g solute / 1e6 g solution) * 1010 g solution = (1.53e3 * 1010) / 1e6 g = (1.53 * 1.01) g = 1.5453 g. Then moles = 1.5453 / 58.44 ≈ 0.0264 mol. But that seems low. Wait, no, 1530 ppm is 0.153% by mass. Let's check: 1.5453 g solute in 1010 g solution: (1.5453 / 1010) * 100 ≈ 0.153%, which is 1530 ppm (since 0.153% = 1530 ppm). So that's correct. Then molarity is moles per liter. So 0.0264 mol / 1 L ≈ 0.0264 M? Wait, no, wait, no, I think I flipped the formula. Wait, no, let's take 1 kg of solution. Mass of solution = 1 kg = 1000 g. Volume of solution = 1000 g / 1.01 g/mL ≈ 990.099 mL ≈ 0.9901 L. Mass of solute in 1 kg solution: 1530 mg = 1.53 g. Moles of solute: 1.53 g / 58.44 g/mol ≈ 0.02618 mol. Then molarity = 0.02618 mol / 0.9901 L ≈ 0.0264 M. Wait, same result. So maybe that's correct. Wait, but let's check with another approach. Let's use the formula for molarity from ppm:

Molarity (M) = $\frac{ppm \times density}{1000 \times molar \, mass}$

ppm = 1.53×10³, density = 1.01 g/mL = 1.01 kg/L, molar mass = 58.44 g/mol.

So M = (1.53×10³ × 1.01) / (1000 × 58.44) = (1545.3) / (58440) ≈ 0.0264 M. Wait, but that seems low. Wait, no, 1530 ppm is 1.53 g/L (if density is 1 g/mL), but with density 1.01 g/mL, it's 1.53 g solute in 1010 g solution, so per liter, it's 1.53 g * (1000 mL / 1010 g) * (1 L / 1000 mL) ? No, wait, no. Wait, 1 L of solution has 1010 g. So 1.53 g solute in 1010 g solution is (1.53 / 1010) g solute per g solution. So per liter (1010 g solution), solute is 1.53 g. So moles is 1.53 / 58.44 ≈ 0.0262 mol. So molarity is ≈ 0.0262 M, approximately 0.0264 M. So the molarity is approximately 0.0264 M. Wait, but let's recalculate:

1.53×10³ ppm = 1530 ppm = 1530 mg/kg = 1.53 g/kg.

Density = 1.01 g/mL, so 1 kg of solution has volume = 1000 g / 1.01 g/mL = 990.099 mL = 0.990099 L.

Mass of solute in 1 kg solution: 1.53 g.

Moles of solute: 1.53 g / 58.44 g/mol = 0.02618 mol.

Molarity = 0.02618 mol / 0.990099 L = 0.02644 M ≈ 0.0264 M.

Yes, that's correct.

Answer:

The molarity of the NaCl solution is approximately $\boxed{0.0264 \, M}$ (or more precisely, around 0.0264 mol/L).