Question #98e37

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Question #98e37

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Answer 1 · 2016-10-02T03:04:22

Let the molar mass of the given mono-basic acid of (empirical formula $C_{3} H_{4} O_{3}$ $C_3H_4O_3$ ) is $m g/mol$ $" "m" g/mol"$
If 2.2 g is dissolved in water to produce 1 L solution then the strength of the acid solution will be $= 2.2 g/L = \frac{2.2}{m} M$ $=2.2"g/L"=2.2/mM$

Since the acid given is mono basic one the mole ratio in which the acid and NaOH will react is 1:1.

25 mL 0.5M NaOH $\equiv \frac{25 \times 0.5}{1000} mol N a O H = 0.0125 m o l N a O H$ $equiv(25xx0.5)/1000" mol "NaOH=0.0125molNaOH$

As this 25mL NaOH solution completely neutralizes 1 L acid solution containing $\frac{2.2}{m} mol$ $2.2/m" mol"$ acid we can write

$\frac{2.2}{m} : 0.0125 = 1 : 1$ $2.2/m:0.0125=1:1$

$\Rightarrow m = \frac{2.2}{0.0125} = 176 g/mol$ $=>m=2.2/0.0125=176" g/mol"$

The given empirical formula of acid is $C_{3} H_{4} O_{3}$ $C_3H_4O_3$

Let its molecular formula be ${(C_{3} H_{4} O_{3})}_{n}$ $(C_3H_4O_3)_n$

So by this MF its molar mass becomes

$= 3 \times 12 + 4 \times 1 + 3 \times 16) n = 88 n g/mol$ $=3xx12+4xx1+3xx16)n=88n " g/mol"$

So

$88 n = 176$ $88n=176$

$n = \frac{176}{88} = 2$ $n=176/88=2$

Hence the Molecular formula of the acid is

${(C_{3} H_{4} O_{3})}_{2} = C_{6} H_{8} O_{6}$ $(C_3H_4O_3)_2=C_6H_8O_6$

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Question #98e37

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