Answer:
Compound B has greater molar mass.
Explanation:
The depression in freezing point is given by ;
..[1]
![m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BMass%20of%20solvent%20in%20kg%7D%7D)
Where:
i = van't Hoff factor
= Molal depression constant
m = molality of the solution
According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.
The depression in freezing point of solution with A solute: ![\Delta T_{f,A}](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%2CA%7D)
Molar mass of A = ![M_A](https://tex.z-dn.net/?f=M_A)
The depression in freezing point of solution with B solute: ![\Delta T_{f,B}](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%2CB%7D)
Molar mass of B = ![M_B](https://tex.z-dn.net/?f=M_B)
![\Delta T_{f,A}>\Delta T_{f,B}](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%2CA%7D%3E%5CDelta%20T_%7Bf%2CB%7D)
As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.
![\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}](https://tex.z-dn.net/?f=%5CDelta%20T_f%5Cpropto%20%5Cfrac%7B1%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%7D)
![M_A](https://tex.z-dn.net/?f=M_A%3CM_B)
This means compound B has greater molar mass than compound A,
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
HCl + NaOH = NaCl + H₂O
n(HCl)=c(HCl)v(HCl)
n(NaOH)=c(NaOH)v(NaOH)
n(HCl)=n(NaOH)
c(HCl)v(HCl)=c(NaOH)v(NaOH)
c(HCl)=c(NaOH)v(NaOH)/v(HCl)
c(HCl)=0.100mmol/mL*60.0mL/150mL=0.040 mmol/mL = 0.040 mol/L
0.040 M HCl
Answer: B. Algae are an r-selected species because they have a very rapid growth phase.
Reasoning: r-selected species are those that emphasize high growth rates.