Question

the weight of AgNO3 (Mol.wt = 170) present in 100 ml of its 0.25 M solution is ​

Answer 1

Answer:

Given

molecular weight of AgNo3 =170

volume , v =100ml

molarity , M = w÷M×1000÷v

0.25 =w÷170×10

w =4.25 gm

Explanation:

Answer 2

Answer:

42.5g

Explanation:

Molarity = {Moles solute}/{Vol.of solution(L)}

-

 

moles of AgNO_{3} in 100ml = 0.25x 10^{-1}  

      = 0.025

mass = 0.025x 170g=4.25g