Chem Concentration Calculation

when encountering this kind of question, break it down into shorter sessions.

first, calculate everything you can immediately think of.
no. of mole of nitrate ion in 100ml 2M NaNO3 solution
= volume x (molarity of salt) x (no. of ion in 1 mole of salt)
= (100/1000) x 2.00 x 1
= 0.2 mole

no. of mole of nitrate in 50ml 0.25M Zn(NO3)2 sol.
= volume x molarity x no. of ion
= (50/1000) x 0.25 x 2
= 0.025 mole


next, think of the question. it's mixing two solutions;
thus final volume of mixture = sum of volumes of all solutions
= 100ml + 50ml = 150ml

ultimately, final concentration
= final no. of mole / final volume
= (0.2 + 0.025) / (150/1000)
= 1.5 M