Answer:- 171 g
Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.
We know that molarity is moles of solute per liter of solution.
If molarity and volume is given then, moles of solute is molarity times volume in liters.
moles of solute = molarity* liters of solution
moles of solute = 0.5*1 = 0.5 moles
To convert the moles to grams we multiply the moles by molar mass.
Molar mass of sucrose = 12(12) + 22(1) + 11(16)
= 144 + 22 + 176
= 342 grams per mol
grams of sucrose required = moles * molar mass
grams of sucrose required = 0.5*342 = 171 g
So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.
5cm
Explanation:
you have to divide 100cm by 20 sec
Answer:
I) 0.0585 M
ii)6.2 g dm-3
Explanation:
The reaction equation is given as;
Na2CO3(aq) +2HCl(aq)------> 2NaCl(aq) + CO2(g) +H2O(l)
Concentration of acid CA= 0.3 M
Volume of acid VA= 3.9 cm^3
Concentration of base CB= the unknown
Volume of base VB= 10 cm^3
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB=CBVBNA
CB= CAVANB/VBNA
substituting values;
CB= 0.3 × 3.9 × 1/ 10.0 × 2
CB= 0.0585 M
ii) mass concentration= molar concentration × molar mass
Molar mass of Na2CO3= 106 gmol-1
Mass concentration= 0.0585 × 106 = 6.2 g dm-3