JEE MAIN - Chemistry (2012 - No. 2)
The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is
1.15 g/mL. The molarity of this solution is :
0.50 M
1.78 M
1.02 M
2.05 M
Explanation
We know molarity (M) = $${{no\,of\,moles\,of\,solute} \over {volume\,of\,solution\,in\,litre}}$$
Moles of solute = $${{120} \over {60}}$$ = 2
Mass of solution = 1000 + 120 = 1120 gm
Density of solution = 1.15 g/mL
$$\therefore$$ volume of solution = $${{1120} \over {1.15}}$$ = 973.9 mL = $${{973.9} \over {1000}}$$ litre = 0.9739 litre
$$\therefore$$ M = $${2 \over {0.9739}}$$ = 2.05
Moles of solute = $${{120} \over {60}}$$ = 2
Mass of solution = 1000 + 120 = 1120 gm
Density of solution = 1.15 g/mL
$$\therefore$$ volume of solution = $${{1120} \over {1.15}}$$ = 973.9 mL = $${{973.9} \over {1000}}$$ litre = 0.9739 litre
$$\therefore$$ M = $${2 \over {0.9739}}$$ = 2.05
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