What volume (in mL) of gaseous NH_3 at N.T.P. | vedclass

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(C) Initial milliequivalents of $H_2SO_4 = N_1 	imes V_1 = 1 \, N 	imes 30 \, mL = 30 \, meq$.Final milliequivalents of $H_2SO_4 = N_2 	imes V_2 =

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$537.6$

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(C) Initial milliequivalents of $H_2SO_4 = N_1 	imes V_1 = 1 \, N 	imes 30 \, mL = 30 \, meq$.Final milliequivalents of $H_2SO_4 = N_2 	imes V_2 = 0.2 \, N 	imes 30 \, mL = 6 \, meq$.Milliequivalents of $NH_3$ required to neutralize the acid = $30 - 6 = 24 \, meq$.Since $NH_3$ is a base,$1 \, meq$ of $NH_3$ corresponds to $1 \, mmol$ of $NH_3$.Therefore,$24 \, mmol$ of $NH_3$ is required.At $N.T.P.$,$1 \, mole$ of gas occupies $22400 \, mL$.Volume of $NH_3 = 24 	imes 10^{-3} \, mol 	imes 22400 \, mL/mol = 537.6 \, mL$.

What volume (in $mL$) of gaseous $NH_3$ at $N.T.P.$ is required to be passed into $30 \, mL$ of $1 \, N \, H_2SO_4$ solution to reduce the acid strength of this solution to $0.2 \, N$?

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