Calculate the molarity of each of the following solutions

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Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

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\textbf{molarity is the number of mole per }\\\textbf{unit volume in Litre}

(a) weight of Co(NO3)2.6H2O = 30g
molar mass of Co(NO3)2.6H2O = 291 g/mol.
volume of solution = 4.3 L
molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}
so, number of moles of solute = given weight/molar mass
= 30/291 = 0.103 mol

molarity = 0.103mol/4.3L = 0.023M

(b) Given,
30 mL of 0.5 M H2SO4 diluted to 500 mL.
In 1000 mL of 0.5 M H2SO4, number of moles present is 0.5 mol.
∴ In 30 mL of 0.5 M H2SO4, number of moles present = 30 × 0.5/1000 = 0.015 mol
∴ molarity = number of moles present/volume
= 0.015mol/volume of solution
= 0.015/0.5L
= 0.03M

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