Question

Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO3 and water. The volume of the solution is 1.421 liters. What is the mass of solute in the solution? Express your answer to three significant figures.

Answer 1

Answer : The mass of solute in solution is $1.09\times 10^2g$.

Solution : Given,

Molarity = 0.730 M

Volume of solution = 1.421 L

Molar mass of sodium carbonate = 105.98 g/mole

Formula used for Molarity :

$Molarity=\frac{w}{M\times V}$

where,

w = mass of solute

M = Molar mass of solute

V = volume of solution in liter

Sodium carbonate is solute and water is solvent.

Now put the given values in above formula, we get the mass of solute in solution.

$0.730mole/L=\frac{w}{(105.98g/mole)\times (1.412L)}$

By rearranging the terms, we get

$m=109.247g=1.09\times 10^2g$

Therefore, the mass of solute in solution is $1.09\times 10^2g$.

Answer 2

The mass of the NaCO₃ solute in the solution is 109.93 grams.

What is molarity?

Molarity is define as the number of moles of solute present in per liter of solution i.e.

M = n/V

From these equation first we calculate the moles of sodium carbonate in 1.421L of solution, whose molarity is 0.730 as:

Moles of NaCO₃ = 0.730 × 1.421 = 1.037 moles

Now we calculate mass of NaCO₃ by using the 1.037 moles as follow:

n = W/M, where

M = molar mass of NaCO₃ = 105.98 g/mole

W = required mass ?

W = 1.037mol × 105.98g/mol = 109.93 grams

Hence, 109.93 grams is the mass of solute.

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