Answer:
OPTION C is correct
C) 1.07 g
Explanation:
CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)3(aq)
But we know molarity molarity= number of moles of solute/ volume of the solution
M= n/V
From the equation above
number of moles of Cacl2 = (37.5 ×0 .100 × 10^-3) = 0.00375 moles.
Then
1 mole of Cacl2 yields 2 moles of Agcl2
0.00375 moles of Cacl2 will produce let say Y.
Y= (0.00375 ×2)/1
= 0.0075 moles.
Number of moles of Agcl2 = mass /molar mass of Agcl
Molar mass of Agcl = 178
Then mass = 178 ×0.0075 = 1.047
Therefore, the mass of agcl precipitate is
1.07 g
