Question

Answer 1

a) Giả sử n _NaNO3 = 1(mol)

=> \(m_{ct\left(NaNO_3\right)}=1.85=85\left(g\right)\)

\(m_{ddNaNO_3\left(40\%0\%\right)}=\dfrac{85}{40\%}=212,5\left(g\right)\)

=> \(m_{H_2O}=212,5-85=127,5\left(g\right)\)

\(S=\dfrac{m_{ct}}{m_{dd}}.100=\dfrac{85}{127,5}.100=66,67\left(g\right)\)

Answer 2

b) \(n_{H_2SO_4\left(4M\right)}=4.4=16\left(mol\right)\)

\(m_{ddH_2SO_4\left(98\%\right)}=\dfrac{16.98}{98\%}=1600\left(g\right)\)

=> \(V_{ddH_2SO_4}=\dfrac{1600}{1,84}=869,57\left(ml\right)\)