\(V_2 = ?\)
\({N_1}{V_1} = {N_2}{V_2}\)
or \(10 \times 10 = 0.1 \times {V_2}\)
or \({V_2} = \frac{{10 \times 10}}{{0.1}},\,{V_2} = 1000\,ml\)
Volume of water to be added
\( = {V_2} - {V_1} = 1000 - 10 = 990\,ml\)
[અણુ દળ - ${Na}: 23.0\, {u}, {O}: 16.0\, {u}, {P}: 31.0 \,{u}]$