Question

how much NaNO3 must be weighed out to make 50ml of an aqueous solution containing 70mg Na ions per ml

Answer 1

Here, Concentration of Na+ ion = 70 mg/ml

So, for 50 mL of solution, mass of Na+ ion = 70 x50 =3500 mg = 3500 x 10-3 gm = 3.5 gm

Now molar mass of NaNO3 = 23 + 14 + 48 = 85 gm

So, %age composition of Na, in NaNO3 = (23/85 )x 100= 27%

Hence, we see that, 100 gm of NaNO3 contains = 27 gm of Na+ ion

Thus mass of NaNO3 containing 3.5 gm Na+ ion = (100/27) x 3.5 =12.96 gm (nearly 13 gm)

Answer 2

The electronic concentration in the hydrogen atom is given by Rydberg equation: del E = - Rh((1/nf^2)-(1/ni^2)).

The emission spectrum of atomic hydrogen is divided into numerous spectral series with wavelengths given by Rydberg formula.

These observed spectral lines are due to the electron making transitions between two energy levels in the atom.