Question

What is the molarity of a solution that contains 85.0 grams of Na2SO4 in 325 milliliters of solution? (The mass of one mole of Na2SO4 is 142 grams.)
0.195 M
0.599 M
1.84 M
6.22 M

Answer 1

First find the number of mol
85/142 = 0.598 mol

divide the mol by the volume in L
0.598/0.325 = 1.81M. C is close enough

Answer 2

Answer : The molarity of a solution is, 1.84 M

Explanation : Given,

Mass of $Na_2SO_4$ = 85.0 g

Molar mass of $Na_2SO_4$ = 142 g/mole

Volume of solution = 325 ml

Molarity : It is defined as the number of moles of solute present in one liter of solution.

In this question, the solute is $Na_2SO_4$.

Formula used :

$Molarity=\frac{w_b}{M_b\times V}\times 1000$

where,

$w_b$ = mass of solute $Na_2SO_4$

$M_b$ = molar mass of solute $Na_2SO_4$

$V$ = volume of solution in ml

Now put all the given values in the above formula, we get the molarity of the solution.

$Molarity=\frac{85.0g}{142g/mole\times 325ml}\times 1000=1.84mole/L=1.84M$

Therefore, the molarity of the solution is, 1.84 M