Respuesta :
Answer:
6.05 g
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
From the question ,
M = 200mM
Since,
1 mM = 10⁻³ M
M = 200 * 10⁻³ M
V = 250 mL
Since,
1 mL = 10⁻³ L
V = 250 * 10⁻³ L
The moles can be calculated , by using the above relation,
M = n / V
Putting the respective values ,
200 * 10⁻³ M = n / 250 * 10⁻³ L
n = 0.05 mol
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
m = 121 g/mol
n = 0.05 mol ( calculated above )
The mass of tri base can be calculated by using the above equation ,
n = w / m
Putting the respective values ,
0.05 mol = w / 121 g/mol
w = 0.05 mol * 121 g/mol
w = 6.05 g
Answer:
We need 6.05 grams of tris base to make this solution
Explanation:
Step 1: Data given
Molecular weight of tris base = 121 g/mol
volume of solution = 250 mL = 0.250 L
Molarity = 200 mM = 0.200 M
Step 2: Calculate moles tris base
Moles tris base = molarity * volume
Moles tris base = 0.200 M * 0.250 L
Moles tris base = 0.05 moles
Step 3: Calculate mass of tris base
Mass tris base = moles tris base * molar mass
Mass tris base = 0.05 moles * 121 g/mol
Mass tris base = 6.05 grams
We need 6.05 grams of tris base to make this solution
