Question

question 20
how many mol of naoh are present in 646 ml of 1 m solution? give your answer to two decimal spaces.
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question 21
how many grams of nacl are required to make 3.5 l of a 0.39 m solution?
g
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question 22
if 0.515 l of a solution was prepared by dissolving 122.7 g of na₃po₄ in 0.489 kg of water...
how many moles of na₃po₄ do you have?
what is the molarity of the solution? m
what is the molality of the solution? m

Explanation:

Question 20

Step1: Recall Molarity Formula

Molarity ($M$) is defined as moles of solute ($n$) per liter of solution ($V$), so $M=\frac{n}{V}$. We need to find $n$, so rearrange to $n = M\times V$.

Step2: Convert Volume to Liters

The volume is 646 mL. Since 1 L = 1000 mL, $V=\frac{646}{1000}=0.646$ L.

Step3: Calculate Moles

Given $M = 1$ M and $V = 0.646$ L, $n=1\times0.646 = 0.646$ mol. Rounded to two decimal places, it's 0.65 mol.

Step1: Recall Molarity Formula

$M=\frac{n}{V}$, so $n = M\times V$. First, find moles of NaCl.

Step2: Calculate Moles of NaCl

Given $M = 0.39$ M and $V = 3.5$ L, $n=0.39\times3.5 = 1.365$ mol.

Step3: Find Molar Mass of NaCl

Molar mass of Na is 22.99 g/mol, Cl is 35.45 g/mol. So molar mass of NaCl is $22.99 + 35.45 = 58.44$ g/mol.

Step4: Calculate Mass

Mass ($m$) = moles ($n$) × molar mass. So $m = 1.365\times58.44\approx79.77$ g.

Step1: Find Molar Mass of $\ce{Na3PO4}$

Molar mass of Na: 22.99 g/mol, P: 30.97 g/mol, O: 16.00 g/mol. So molar mass $= 3\times22.99+30.97 + 4\times16.00=68.97+30.97 + 64.00 = 163.94$ g/mol.

Step2: Calculate Moles

Moles ($n$) = mass ($m$) / molar mass. Given $m = 122.7$ g, $n=\frac{122.7}{163.94}\approx0.748$ mol.

Answer:

0.65

Question 21