Question

The sulphate of a metal M contains 9.87% of M.This sulphate is isomorphous with ZNSO4.7H2O.The atomic weight of M is

Answer 1

Answer:

As the given sulphate is isomorphous with ZnSO4.7H2O its formula would be MSO4.7H2O.m is the atomic weight of M, molecular weight of MSO4.7H2O =m+32+64+126 =m+222                    Hence % of M=mm+222×100=9.87(given)  or                    100m=9.87m+222×9.87or 90.13m=222×9.87                                 or m=222×9.8790.13=24.3.

Explanation:

Answer 2

Answer:

The atomic wt. of $MSO_{4}.7H_{2}O$ is 24.3

Explanation:

As per the data given in the question,

The given sulphate is isomorphous with $ZnSO_{4}.7H_{2}O$.

So, its formula will be $MSO_{4}.7H_{2}O$

Then molar mass of $MSO_{4}.7H_{2}O$ will be = $M+32+64+7\times 18= M+222$

Now, % of M in the compund $MSO_{4}.7H_{2}O$=

$=\frac{(Atomicity)\times(Atomic wt of M)}{(Molar mass of MSO_{4}.7H_{2}O) } \times 100$

$9.87=\frac{1\times M }{M+222} \times 100\\100M=9.84(M+222)\\M=\frac{2191.14}{90.13}\\ M=24.3$

Hence the atomic wt. of $MSO_{4}.7H_{2}O$ is 24.3

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