Answer:
0.221M
Explanation:
From the question ,
The Molarity of AgNO₂ = 0.310 M
Hence , the concentration of Ag⁺ = 0.301 M
The volume of AgNO₂ = 250 mL
and,
The Molarity of Sodium chromate = 0.160 M
The volume of Sodium chromate = 100 mL.
As the solution is mixed the final volume becomes ,
250mL +100mL = 350mL
Now, using the formula , to find the final molarity of the mixture ,
M₁V₁ ( initial ) = M₂V₂ ( final )
substituting the values , in the above equation ,
0.310M * 250ml = M₂ * 350ml
M₂ = 0.221M
Hence , the concentration of the silver in the final solution = 0.221M
Molas mass C6H12O6 = 180.15 g/mol
1 mole C6H12O6 -------------- 180.15 g
?? molesC6H12O6 ------------ 3.10x10⁻³ g
3.10x10⁻³ x 1 / 180.15 => 0.00001720 moles
1 mole ------------------------- 6.02x10²³ molecules
0.00001720 moles ---------- ??
0.00001720 x (6.02x10²³) / 1 => 1.035x10¹⁹ molecules
0.005 is the correct answer i hope this helps man :)
