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1.	.1; पल aneb 1l bs- . Chloand - 9 0100mod K ge gl whak vy 4 un,w" .00 mwlidoy fovmala) ं
Answer» In 100g of sample of the compound, 4.07g of hydrogen, 24.27g of carbon and 71.65g of chlorine are present. Moles of hydrogen= 4.07g/ 1g = 4.0 Moles of carbon= 24.27g/ 12g = 2.0 Moles of chlorine= 71.65g/35g = 2.0 Since 2.0 is the smallest value, so by dividing each of the mole values obtained by this smallest value we will get a ratio of 2:1:1 for H:C:Cl. Thus, the empirical formula of the compound is CH2Cl. For CH2Cl, empirical formula mass= 12+ (2×1) +35 = 49g. n = Molar mass/ empirical Formula mass= 98.96g/ 49g = 2 Therefore, Empirical formula = CH2Cl n=2 Hence, molecular formula= C2H4Cl2.

.1; पल aneb 1l bs- . Chloand - 9 0100mod K ge gl whak vy 4 un,w" .00 mwlidoy fovmala) ं

Answer»

In 100g of sample of the compound, 4.07g of hydrogen, 24.27g of carbon and 71.65g of chlorine are present.

Moles of hydrogen= 4.07g/ 1g = 4.0 Moles of carbon= 24.27g/ 12g = 2.0 Moles of chlorine= 71.65g/35g = 2.0

Since 2.0 is the smallest value, so by dividing each of the mole values obtained by this smallest value we will get a ratio of 2:1:1 for H:C:Cl.

Thus, the empirical formula of the compound is CH2Cl.

For CH2Cl, empirical formula mass= 12+ (2×1) +35 = 49g. n = Molar mass/ empirical Formula mass= 98.96g/ 49g = 2

Therefore, Empirical formula = CH2Cl n=2

Hence, molecular formula= C2H4Cl2.