Question

Calculate the percentage of water of crystallization in 10NaCOHO
[Na – 23, C – 12, O – 16, H – 1]

Answer 1

${\huge{\underline{\underline{\mathrm{Correct~QuestioN::}}}}}$

Calculate the percentage of water of crystallization in 10NaHCO3

[Na – 23, C – 12, O – 16, H – 1]

${\huge{\underline{\underline{\mathrm{AnsweR::}}}}}$

ㅤㅤㅤㅤㅤㅤA/Q::

• Na = 23×10 = 230
• H = 1×10 = 10
• C = 12×10 = 120
• O = 16×30 = 480

ㅤㅤㅤㅤㅤTherefore,

Total , R.A.M (Relative Atomic Mass) = 230+10+120+480

= 840 a.m.u

ㅤㅤㅤㅤㅤㅤㅤNow;

R.A.M of water = 10(H2O) = 10( 1×2+16×1)

= 10( 2+16) = 10×18 = 180 a.m.u

So, Percentage

= (Weight of the molecule/Total weight of the compound) × 100

$\tt\implies$ (Weight of water/Total weight of Sodium Bicarbonate) × 100

$\tt\implies$ (180/840) × 100 = 21.42 %

Answer 2

Answer:

Calculate the percentage of water of crystallization in 10NaHCO3

[Na – 23, C – 12, O – 16, H – 1]

{\huge{\underline{\underline{\mathrm{AnsweR::}}}}}AnsweR::

ㅤㅤㅤㅤㅤㅤA/Q::Na = 23×10 = 230H = 1×10 = 10C = 12×10 = 120O = 16×30 = 480ㅤㅤㅤㅤㅤTherefore,

Total , R.A.M (Relative Atomic Mass) = 230+10+120+480

= 840 a.m.u

ㅤㅤㅤㅤㅤㅤㅤNow;

R.A.M of water = 10(H2O) = 10( 1×2+16×1)

= 10( 2+16) = 10×18 = 180 a.m.u

So, Percentage

= (Weight of the molecule/Total weight of the compound) × 100

\tt\implies⟹ (Weight of water/Total weight of Sodium Bicarbonate) × 100

\tt\implies⟹ (180/840) × 100 = 21.42 %