Calculate the volume in milliliters of a sodium carbonate solution that contains of sodium carbonate. Be sure your answer has the correct number of significant digits.

25.0 g of sodium carbonate are present in 220 ml of the solution.

Explanation:

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Calculate the volume in milliliters of a 1.07 M sodium carbonate solution that contains 25.0g of sodium carbonate Na2CO3. Be sure your answer has the correct number of significant digits.

First, let's find how many moles of sodium carbonate have a mass of 25.0 g.

The molar mass of Na₂CO₃ is 106 g/mol.

So, if 106 g of sodium carbonate is equal to 1 mol, then 25.0 g will be:

25.0 g · (1 mol / 106 g) = 0.236 mol Na₂CO₃

The solution contains 1.07 mol sodium carbonate in 1 liter.

So, if 1.07 mol sodium carbonate is present in 1 l solution, then, 0.236 mol will be present in:

0.236 mol · (1000 ml / 1.07 mo) = 221 ml (220 ml without any intermediate rounding).

25.0 g of sodium carbonate are present in 220 ml of the solution.