A certain current liberates 0.5 g of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?

12.7 g

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

15. 9 g

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

31.8 g

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

63. 5 g

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

Open in App

Solution

The correct option is C 15. 9 g
$\frac{W e i g h t o f H_{2}}{E w o f H_{2}} = \frac{W e i g h t o f C u}{E w o f C u}$
$\frac{0.5}{1} = \frac{x}{63.5 / 2}$
$∴ x = 15.875 g \approx 15.9 g$

Hence, the correct option is $B$

Suggest Corrections

Similar questions

Q. A certain amount of current liberates

0.5 g

of hydrogen in 2 hours. How many grams of copper can be produced by the same current flowing for the same time in copper sulphate solution?

A certain current liberated 0.504 gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution

A certain current liberated $0.504$ g of hydrogen in $2$ hrs. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?

A certain current liberated 0.504 gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution

Q. A certain current liberated

0.504 g

of hydrogen in

2

hours. How many grams of copper can be liberated by the same current flowing for the same time in

C u S O_{4}

solution?

Join BYJU'S Learning Program

A certain current liberates 0.5 g of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?

The correct option is C 15. 9 gWeightofH2EwofH2=WeightofCuEwofCu0.51=x63.5/2∴x=15.875g≈15.9 gHence, the correct option is B

The correct option is C 15. 9 g
$\frac{W e i g h t o f H_{2}}{E w o f H_{2}} = \frac{W e i g h t o f C u}{E w o f C u}$
$\frac{0.5}{1} = \frac{x}{63.5 / 2}$
$∴ x = 15.875 g \approx 15.9 g$

Hence, the correct option is $B$