<span>homogeneous because particles are mixed uniformly</span>
Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
1) Write the balanced chemical equation
2HCl + Na2 CO3 ----------> 2NaCl + H2CO3
2) Write the molar ratios:
2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3
3) Convert 0.15g of sodium carbonate to number of moles
3a) Calculate the molar mass of Na2CO3
Na: 2 * 23 g/mol = 46 g/mol
C: 12 g/mol =
O: 3 * 16 g/mol = 48 g/mol
molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol
3b.- Calculate the number of moles of Na2CO3
# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3
4) Calculate the number of moles of HCl from the molar proportion:
[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl
5) Calculate the volume of HCl from the definition of Molarity
Molarity, M = # moles / volume in liters
=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters
0.028 liters * 1000 ml / liter = 28 ml.
Answer: 28 mililiters of 0.1 M HCl.
