Answer:
10.23 grams of sucrose should be added.
Explanation:
1.15 m means molality (moles of solute in 1kg of solvent)
1.15 moles of sucrose are contained in 1 kg of solvent (1000 g)
Let's determine the moles in our mass of solvent.
Firstly we convert the g to kg → 26 g . 1kg/1000g = 0.026 kg
m . mass (kg) = 1.15 mol/kg . 0.026kg → 0.0299 moles.
Finally we convert the moles to mass (mol . molar mass)
0.0299 mol . 342.3 g/mol = 10.23 g
173.1f is the answer I believe, please let me know if I'm wrong then I would try to make up for it
The complete table is inserted.
A table is given,
Formulas used:
pH= -log(H⁺)
pOH= -log(OH⁻)
pH+ pOH=14
Calculations:
For A: (H⁺)=2×10⁻⁸M
Using the pH formula:
pH= -log(H⁺)=-log(2×10⁻⁸)=7.69
pOH=14 - 7.69=6.3
Calculating OH concentration,
pOH= -log(OH⁻)
6.3= -log(OH⁻)
(OH⁻)=5.011×10⁻⁷M
Hence, the nature of A is basic.
Similarily,
For B,
(OH⁻)=1×10⁻⁷
Using the pH formula:
pOH= -log(OH⁻)= -log(1×10⁻⁷)=7
pH=14-7=7
Calculating H concentration,
pH= -log(H⁺)
7= -log(H⁺)
(H⁺)=1×10⁻⁷M
Hence, the nature of B is neutral.
Similarily,
For C,
pH=12.3
Using the pH formula:
pOH=14-12.3=1.7
Calculating H concentration,
pH= -log(H⁺)
12.3= -log(H⁺)
(H⁺)=5.011×10⁻¹³M
Calculating OH concentration,
pOH= -log(OH⁻)
1.7= -log(OH⁻)
(OH⁻)=1.99×10⁻²M
Hence, the nature of C is Basic.
Similarily,
For D,
pOH=6.8
Using the pH formula:
pH=14-6.8=7.2
Calculating H concentration,
pH= -log(H⁺)
7.2= -log(H⁺)
(H⁺)=6.309×10⁻⁸M
Calculating OH concentration,
pOH= -log(OH⁻)
6.8= -log(OH⁻)
(OH⁻)=1.58×10⁻⁷M
Hence, the nature of D is basic.
Learn more about the acid and bases here:
brainly.com/question/16189013
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The electron configuration for Sodium (Na) is [Ne] 3s1