Question

Need an answer urgently please!!
Q.) 2.34g pure sample of MgSO4.7H2O is heated to

evaporate all of its water content. Maximum mass loss of

the sample will be: -

1. 1.8g 2. 0.7g

3. 1.26g. 4. 0.36g​

Answer 1

Explanation:

given=

given weight=2.34g

molecular weight of MgSO4.7H2O--246g

for 246g---------120g(MgSO4)

for 2.34g-----------x

by calculating wegot x=1.12

7h2o= 2.34-1.12=1.2

so the mass loss is ~1.26 g

plss mark me as brainlist

Answer 2

The maximum mass loss of sample will be 3. 1.26g.

Given - Mass of sample

Find - Mass of loss sample

Solution - The molar mass of $MgSO_{4}. 7H_{2}O$ is 246.48 grams.

As we know, the molar mass of water is 18 grams.

So, the molar mass of heptahydrate will be : 18 × 7

Molar mass = 126 grams

Mass of $MgSO_{4}$ will be = 246.48 - 126

Mass of $MgSO_{4}$ = 120.48 grams

Now, 246.48 grams contains 120.48 grams of $MgSO_{4}$

So, 2.34 grams of sample will have amount of $MgSO_{4}$ = $\frac{120.48}{246.48} \times 2.34$

Amount of $MgSO_{4}$ in given sample = 1.14 grams

Amount of $MgSO_{4}$ = 2.34 - 1.14

Amount = 1.2 grams

Hence, the mass of water loss on evaporation will be 3. 1.26 grams.

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