1) 1M SOLUTION OF NaNO₃ HAS DENSITY 1.25 g cm^-3. Calculate molality.

2) how many grams of KCLO₃ MUST BE DECOMPOSED TO PREPARE 3.36 LITRES OF OXYGEN AT STP

Can u explain it briefly?

I hope this is useful

Given that,  1M solution of NaNO3,
By definition if Molarity we know that it is the no. of moles present in per litre of solution. => 1M molar solution has 1 mole of NaNO3 for 1000ml of solution.
Therefore, mass of solute is mass of 1 mole of NaNO3 = 23+14+48 = 85 gm.
                  volume of solution = 1000 ml as per definition.

Now,Given that density is 1.25 g/cm3
Therefore, mass = 1.25 * 1000 (m = density*volume)
=> mass = 1250 gm (solution mass)
Mass of solvent = mass of solution - mass of solute
                          = 1250 - 85= 1165gm.

Molarity is the no. of mole of solute per kg of solvent
=> m = (1/1165) * 1000 = 1000/1165 = 0.85 m (approx)

Hope ot helps

Hope it will best answer.

While calculating molality in last step why have we multiplied it by 1000