1 g of hydrogen is found to combine with 80 g of bromine. 1 g of calcium combines with 4 g of bromine. The equivalent mass of calcium is:

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Solution

The correct option is B 20
$\frac{1}{2}$ $H_{2}$ + $\frac{1}{2}$ $B r_{2}$ $\to$ HBr

Ca + $B r_{2}$ $\to$ $C a B r_{2}$
4g of broime react with 1 g of Ca
160 g of bromine react with = $\frac{1 * 160}{4}$ = 40 g

equivalent mass of calcium = $\frac{a t o m i c m a s s}{v a l a n c y}$

equivalent mass of calcium = $\frac{40}{2}$

equivalent mass of calcium = 20

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Similar questions

Q. One gram of hydrogen is found to combine with 80 g of bromine. One gram of calcium (valency = 2) combines with 4 g of bromine. The equivalent weight of calcium is :

Q. One gram of hydrogen is found to combine with

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Q. 1.0 g of a metal combines with 8.89 g of Bromine. Equivalent weight of metal is nearly :

KBr contains 32.9% by weight of potassium. If 6.40 g of bromine reacts with 3.60 g of potassium, calculate the number of moles of potassium which combine with bromine to form KBr.

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1 g of hydrogen is found to combine with 80 g of bromine. 1 g of calcium combines with 4 g of bromine. The equivalent mass of calcium is:

The correct option is B 2012H2 + 12Br2 → HBrCa + Br2 → CaBr24g of broime react with 1 g of Ca160 g of bromine react with = 1∗1604 = 40 gequivalent mass of calcium = atomicmassvalancyequivalent mass of calcium = 402equivalent mass of calcium = 20