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NCERT Class XII Chemistry
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Question : 44 of 53
Marks: +1, -0
Calculate the molarity of each of the following solutions :
(a) 30 g of Co(NO3)2⋅6H2O\mathrm{Co(NO_3)_2 \cdot 6H_2O} in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4\mathrm{H_2SO_4} diluted to 500 mL (atomic mass of cobalt = 58.7)
Solution:  
(a) Molar mass of Co(NO3)2⋅6H2O\text{(a) Molar mass of } \mathrm{Co(NO_3)_2 \cdot 6H_2O} =58.7+2(14+48)+6×18=58.7+2(14+48)+6 \times 18
=290.7g mol−1=290.7 \mathrm{g\,mol}^{-1}
Number of moles of Co(NO3)2⋅6H2O=MassMolar mass\mathrm{Co(NO_3)_2 \cdot 6H_2O} = \frac{\text{Mass}}{\text{Molar mass}}
=30 g290.7 g mol−1=0.103= \frac{30 \,\mathrm{g}}{290.7 \,\mathrm{g\,mol}^{-1}} = 0.103
Volume of solution =4.3 L=4.3 \,\mathrm{L}
Molarity of solution =Number of moles of soluteVolume of solution in L= \frac{\text{Number of moles of solute}}{\text{Volume of solution in } \mathrm{L}}
=0.103 mole4.3 L=0.024 M= \frac{0.103 \,\text{mole}}{4.3 \,\mathrm{L}} = 0.024 \,\mathrm{M}
(b) Given V1=30 mL,V_{1}=30 \,\mathrm{mL}, M1=0.5,V2=500 mL,M_{1}=0.5, V_{2}=500 \,\mathrm{mL}, M2=?M_{2}=?
Using formula, V1M1=V2M2V_{1} M_{1}=V_{2} M_{2}
or, 30×0.5=500×M230 \times 0.5=500 \times M_{2}
or, M2=0.03M_{2}=0.03
Molarity of solution =0.03M= 0.03 \mathrm{M}
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