The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity if this solution is

Question

The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity if this solution is (A) 0.50 M (B) 1.78 M (C) 1.02 M (D) 2.05 M

Tardigrade Admin · Accepted Answer

Total mass of solution = 1000 g water + 120 g urea = 1120 g Density of solution =1.15 g/mL Thus, volume of solution =( textmass/ textdensity) = (1120 g/1.15 g/mL) = 973.91 mL = 0.974 L Moles of solute = (120/60) = 2 mol Molarity = ( textmoles of solute/ textvolume(L)of solution) = (2 mol/0.974 L) = 2.05 mol L-1

Q. The density of a solution prepared by dissolving $120 g$ of urea (mol. Mass $= 60 u$ ) in $1000 g$ of water is $1.15 g / m L$ . The molarity if this solution is

$0.50 M$

$1.78 M$

$1.02 M$

$2.05 M$

Q. The density of a solution prepared by dissolving 120g of urea (mol. Mass =60u) in 1000g of water is 1.15g/mL. The molarity if this solution is

0.50M

1.78M

1.02M

2.05M

Total mass of solution =1000 g water +120 g urea =1120g Density of solution =1.15g/mL Thus, volume of solution =densitymass​=1.15g/mL1120g​ =973.91mL=0.974L Moles of solute =60120​=2mol Molarity =volume(L)of solutionmoles of solute​ =0.974L2mol​=2.05molL−1

Q. The density of a solution prepared by dissolving $120 g$ of urea (mol. Mass $= 60 u$ ) in $1000 g$ of water is $1.15 g / m L$ . The molarity if this solution is

$0.50 M$

$1.78 M$

$1.02 M$

$2.05 M$