The density of a solution prepared by dissolving $120 g$ of urea $(m o l . m a s s = 60)$ in $1000 g$ of water is $1.15 g / m L$ . The molarity of this solution is:

0.50 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.78 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.02 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.05 M

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $2.05 M$
Total mass of solution = 1120g

moles of urea= $\frac{120}{60} = 2$

Density = $\frac{m a s s}{v o l u m e} = \frac{1120}{V}$

Molarity = $\frac{m o l e s}{v o l u m e i n l i t r e} = \frac{2 \times 1.15}{1120} \times 1000 = 2.05 M$

Suggest Corrections

Similar questions

120 g

1000 g

1.15 g / m L

120 g

60 g {m o l}^{- 1}

1000 g

1.15 g / m l

120

60

1000

1.15

Join BYJU'S Learning Program