Calculate the mass percentage of aspirin $(C_{9} H_{8} O_{4})$ in acetonitrile $(C H_{3} C N)$ when $6.5 g$ of $C_{9} H_{8} O_{4}$ is dissolved in $450 g$ of $C H_{3} C N$ .

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Solution

Total mass, i.e $C_{9} H_{8} O_{4}$ and $C H_{3} C N$ have $180$ and $41$ molecular weight respectively
Mass % of $C_{9} H_{8} O_{4} = \frac{\frac{6.5}{180}}{\frac{450}{41} + \frac{6.5}{180}} \times 100 = 0.32$ %

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Question 2.28

Calculate the mass percentage of aspirin $(C_{9} H_{8} O_{4})$ in acetonitrile $(C H_{3} C N)$ when 6.5g of $C_{9} H_{8} O_{4}$ is dissolved in 450 g of $C H_{3} C N$

Q. Aspirin has the formula

C_{9} H_{8} O_{4}

. Number of moles of aspirin in a tablet weighing

500

mg is

x \times 10^{- 5}

. Then, the value of

x

is ________.

C_{7} H_{6} O_{3} S a l i c y l i c a c i d (138.12 g m o l^{- 1}) + C_{4} H_{6} O_{3} A c e t i c a n h y d r i d e (102.09 g m o l^{- 1}) \to C_{9} H_{8} O_{4} A s p i r i n (180.15 g m o l^{- 1}) + C_{2} H_{4} O_{2} a c e t i c a c i d (60.05 g m o l^{- 1})

What is the percentage yield if 0.85 g of aspirin formed in St reaction of .1 g of salicylic acid with excess of acetic anhydride ?

Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n-hexane and n-octane

(ii) I₂ and CCl₄

(iii) NaClO₄ and water

(iv) methanol and acetone

(v) acetonitrile (CH₃CN) and acetone (C₃H₆O).

QUESTION 2.23(i)

Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n - hexane and n - octane

(ii) $I_{2} a n d C C l_{4}$

(iii) $N a C l O_{4} a n d w a t e r$

(iv) Methanol and acetone

(v) Acetonitrile $(C H_{3} C N)$ and acetone $(C_{3} H_{6} O)$

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Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

Total mass, i.e C9H8O4 and CH3CN have 180 and 41 molecular weight respectively Mass % of C9H8O4=6.518045041+6.5180×100=0.32%

Calculate the mass percentage of aspirin $(C_{9} H_{8} O_{4})$ in acetonitrile $(C H_{3} C N)$ when $6.5 g$ of $C_{9} H_{8} O_{4}$ is dissolved in $450 g$ of $C H_{3} C N$ .

Total mass, i.e $C_{9} H_{8} O_{4}$ and $C H_{3} C N$ have $180$ and $41$ molecular weight respectively
Mass % of $C_{9} H_{8} O_{4} = \frac{\frac{6.5}{180}}{\frac{450}{41} + \frac{6.5}{180}} \times 100 = 0.32$ %