Question

How much Ca (NO3)2 in mg must be present in 50 ml of a solution with 2.35 ppm of cas
a) 0.1175
b) 770.8
c) 4.7
d) 0.48

​

Answer 1

Answer:

C.)4.7 Is the answer of your question

Explanation:

Please please please give me the brilliant answer please please

Answer 2

Answer:

0.48 mg

Explanation:

ppm=2.35

ppm=  mass of solute of in mg ÷volume of solution in litres

2.35=  mass of Ca in mg  ÷50×10  ⁻³

mass of Ca in mg  = 0.1175×10⁻³

moles of Ca=  0.1175×10⁻³ ÷ 40 = 0.0029375

moles of Ca= moles of Ca(NO₃)₂   = 0.0029375×10 ⁻³

Molar mass of  Ca(NO₃)₂  =40+28+96=164 grams

∴  mass of Ca(NO₃)₂  = 164×0.0029375×10⁻³

=0.48×10⁻³ grams

=0.48 mg

option (d)