How many moles of glucose (C6H12O6) are in 5.0 liters of a 2.5 M C6H12O6 solution?
1 answer:
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Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
+
⇄
Formation constant Kf
Kf = / ( [
][
] ) = 5.0 × 10¹⁰
Now,
[] =
; ∝₄ = 0.35
so the equilibrium is;
+
⇄
+ 4H⁺
Given that; =
{ 1 mol
reacts with 1 mol
}
so at equilibrium, =
= x
∴
+
⇄
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = / ( [
][
] ) =
/ ( [
][
] ) = 5.0 × 10¹⁰
⇒ / ( [
][
] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Answer:
0.136g
Explanation:
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
Initial mole of Co(NO3)2
Mole of Co(NO3)2 in final solution
Mole of NO3- in final solution = 2 x Mole of Co(NO3)2
Mass of NO3- in final solution is mole x Molar mass of NO3