The enthalpy of combustion of glucose (mol. wt: 180 g mol-1) is -2840 kJ mol-1. Then the amount of heat evolved when 0.9 g of glucose is burnt, will be

Question

The enthalpy of combustion of glucose (mol. wt: 180 g mol-1) is -2840 kJ mol-1. Then the amount of heat evolved when 0.9 g of glucose is burnt, will be (A) 14.2 kJ (B) 14.2 J (C) 28.4 kJ (D) 1420 kJ

Tardigrade Admin · Accepted Answer

The enthalpy of combustion of glucose (mol.

wt . = 180 g / m o l) = - 2840 k J / m o l

180 g

glucose gives enthalpy of combustion

= 2840 k J / m o l

1g glucose gives enthalpy of combustion

= \frac{2840}{180}

So,

0.9 g

glucose gives enthalpy of combustion

= \frac{2840}{180} \times 0.9

= \frac{2840}{200} k J = 14.2 k J

Q. The enthalpy of combustion of glucose (mol. wt: $180 g m o l^{- 1}$ ) is $- 2840 k J m o l^{- 1}$ . Then the amount of heat evolved when $0.9 g$ of glucose is burnt, will be

$14.2 k J$

$14.2 J$

$28.4 k J$

$1420 k J$

$142 k J$

Q. The enthalpy of combustion of glucose (mol. wt: 180gmol−1) is −2840kJmol−1. Then the amount of heat evolved when 0.9g of glucose is burnt, will be

14.2kJ

14.2J

28.4kJ

1420kJ

142kJ