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How many grams of solute are in each of the following solutions? 0.437 L of a...

How many grams of solute are in each of the following solutions?

0.437 L of a 0.350 M K2CO3 solution =

15.5 mL of a 2.50 M AgNO3 solution =

25.2 mL of a 6.00 M H3PO4 solution =

Solutions

Expert Solution

Molarity of K2CO3 = Number of moles of K2CO3 / Volume of solution in L

0.350 = Number of moles of k2CO3 / 0.437 L

Number of moles of K2CO3 = 0.350 * 0.437 = 0.1530 mole

Number of moles of K2CO3 = mass of K2CO3 / molar mass of K2CO3

0.1530 mole = mass of K2CO3 / 138.205 g.mol^-1

Mass of K2CO3 = 0.1530 * 138.205 = 21.15 g

queen_honey_blossom answered 2 years ago
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