Solveeit Logo
Login

Question

Chemistry Question on Solutions

The volume of water to be added to 100 cm3100\, cm^{3} of 0.5 N H2SO40.5 \,N\, H_{2}SO_{4} to get decinormal concentration is

A

400 cm3400 \,cm^{3}

B

450 cm3450 \,cm^{3}

C

500 cm3500 \,cm^{3}

D

100 cm3100 \,cm^{3}

Answer

400 cm3400 \,cm^{3}

Explanation

Solution

N1V1=N2V2N_{1}V_{1}=N_{2}V_{2} or, 0.5×100=0.1×V20.5\times 100=0.1\times V_{2} or, V2=0.5×1000.1=500 cm3V_{2}=\frac{0.5\times100}{0.1}=500\,cm^{3} ∴(500−100)=400 cm3\therefore (500-100)=400\,cm^{3} water is to be added to 100 cm3100\,cm^{3} of 0.5 N H2SO40.5\,N\,H_{2}SO_{4}