Question

Calculate the volume, in milliliters, for each of the following that provides a given amount of solute.

A) 14.2g of Na2CO3 from a 0.140 M Na2CO3 solution

B) 0.920 mole of NaNO3 from a 0.710 M NaNO3 solution

C) 38.0 g of LiOH from a 2.20 M LiOH solution

Express answers to three significant figures and include appropriate units

Answer 1

we need molar mass of
Na2CO3 = [2*22.99 + 1*12.01 + 3*16] g/mol = 105.99 g/mol
given mass =14. 2 g
number of moles present = 14.2 g/105.99 g/mol = 0.13397 mols
molarity of solution = 0.140 M
volume required = number of moles present /molarity = 0.13397 mols/0.14 M =0.95693 L
volume in ml = 957. ml
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0.920 mole of NaNO3
molarity =0.710 M
volume required = number of moles present /molarity = 0.920mols/0.710 M = 1.2957 L

volume in ml = 130.ml
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molar mass of liOH = [1*6.94 + 1*16+ 1*1.01]g/mol= 23.95 g/mol
mass of LiOH =38.0 g
moles of LiOH present = 38.0 g /23.95 g/mol=1.58664 mols
volume required = number of moles present /molarity =1.58664 mols/2.20 M
=0.7212 L
volume in ml = 721. ml
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all solved .Thank you