Question

how many grams of hcl are needed to make 200.0 ml of a 3.0 m solution?
hcl: 36.46 g/mol
? g

Explanation:

Step1: Recall Molarity Formula

Molarity ($M$) is defined as moles of solute ($n$) divided by volume of solution in liters ($V$). The formula is $M = \frac{n}{V}$. We need to find moles first, so rearrange to $n = M \times V$.
The volume is 200.0 mL, which is $200.0 \times 10^{-3} \text{ L} = 0.2000 \text{ L}$. Molarity is 3.0 M. So $n = 3.0 \text{ mol/L} \times 0.2000 \text{ L} = 0.60 \text{ mol}$.

Step2: Calculate Mass from Moles

Mass ($m$) is moles ($n$) times molar mass ($MM$). The formula is $m = n \times MM$. Molar mass of HCl is 36.46 g/mol. So $m = 0.60 \text{ mol} \times 36.46 \text{ g/mol} = 21.876 \text{ g}$. Rounding appropriately (depending on significant figures, 3.0 has two, 200.0 has four, 36.46 has four; so result can be 22 g or 21.9 g, but let's check calculation again. Wait, 3.0 0.2000 = 0.6000 mol (since 200.0 mL is four sig figs, 3.0 is two, so maybe two sig figs? Wait, 3.0 M is two decimal? No, 3.0 has two significant figures. 200.0 has four. Molar mass 36.46 has four. So when multiplying, the least number of sig figs is two? Wait, 3.0 is two, 0.2000 is four (because the trailing zero after decimal is significant). Wait, 200.0 mL is 0.2000 L, which is four significant figures. 3.0 M is two. So the moles would be 3.0 0.2000 = 0.60 mol (two significant figures? Wait, 3.0 has two, 0.2000 has four, so the product should have two? Wait, no: 3.0 is two sig figs, 0.2000 is four. When multiplying, the result should have the same number of sig figs as the least precise measurement. So 3.0 (two) 0.2000 (four) = 0.60 (two sig figs? Wait, 3.0 0.2000 = 0.6000, but we round to two sig figs? Wait, no, 3.0 has two, so 0.60 (two decimal places? No, sig figs. 3.0 is two sig figs, so 0.60 (two sig figs). Then mass is 0.60 mol 36.46 g/mol = 21.876, which rounds to 22 g (two sig figs) or 21.9 g (if we consider 3.0 as two, 0.2000 as four, so maybe three? Wait, maybe the problem expects us to use exact values. Let's do the calculation without worrying about sig figs first. 3.0 M 0.2000 L = 0.6000 mol. Then 0.6000 mol * 36.46 g/mol = 21.876 g, which is approximately 21.9 g or 22 g. But let's check the steps again.

Wait, step 1: Convert volume to liters. 200.0 mL = 0.2000 L (correct, because 1 L = 1000 mL, so divide by 1000). Step 2: Moles = Molarity Volume = 3.0 mol/L 0.2000 L = 0.6000 mol (since 3.0 has two decimal? No, 3.0 is two significant figures. Wait, 3.0 means two sig figs. 0.2000 is four. So when multiplying, the result should have two sig figs? But 3.0 0.2000 = 0.60 (two sig figs). Then mass: 0.60 36.46 = 21.876, which is 22 g (two sig figs) or 21.9 g (if we take 0.60 as two decimal places, but no, sig figs. Alternatively, maybe the problem doesn't care about sig figs and just wants the numerical answer. Let's recalculate: 3.0 M 0.2000 L = 0.6000 mol. 0.6000 mol 36.46 g/mol = 21.876 g, which is approximately 21.9 g or 22 g. But let's check the exact calculation: 3.0 200.0 / 1000 36.46. 200.0 / 1000 = 0.2. 3.0 0.2 = 0.6. 0.6 36.46 = 21.876, so 21.9 g (or 22 g). But maybe the answer is 21.9 g or 22 g. Wait, let's do it precisely:

Volume $V = 200.0 \text{ mL} = 0.2000 \text{ L}$

Molarity $M = 3.0 \text{ mol/L}$

Moles $n = M \times V = 3.0 \times 0.2000 = 0.6000 \text{ mol}$

Molar mass $MM = 36.46 \text{ g/mol}$

Mass $m = n \times MM = 0.6000 \times 36.46 = 21.876 \text{ g} \approx 21.9 \text{ g}$ (or 22 g if rounding to two significant figures).

Answer:

21.9 (or 22, depending on significant figures; the precise calculation gives 21.876, which can be rounded to 21.9 or 22)