Question

the hydrated salt Na2CO3.nH2O undergoes 57% loss in mass on heating and becomes anhydrous. find vlue of n?

Answer 1

Solution :

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$\rm{Na_2Co_3.nH_2O \longrightarrow Na_2Co_3 + nH_2O}$

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Let, 100 g of Na2CO3.xH2O be present.

Given % loss in mass on heating = 57%

Mass of salt = [2×Na + C + 3×O]

Mass of salt = [2×23 + 12 +3×16]

Mass of salt = 106 g

And when n moles of water is added, then mass

= 106 + n[1×H + 2×O]

= 106 + 18n

Therefore according to the question,

$=>57 = \dfrac{100n}{106 + 18n} \times 18$

$=> \dfrac{57}{18} = \dfrac{100n}{106 + 18n}$

$=>3.16 = \dfrac{100n}{106 + 18n}$

$=>3.16(106 + 18n) = 100n$

$=>334.96 + 56.88n \: = 100n$

$=>334.96 = 100n - 56.88n$

$=> 334.96 = 43.12n$

$=> \dfrac{334.96}{43.12} = n$

$7.76. = n$

So, n = 8 (approximately)

Hence, the value of n is 8.