$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

What is the mass of one mole of H2o

puteri [66]

Answer:the answer is 18.01528

Explanation:

5 0

3 years ago

Write a balanced chemical equation for the complete combustion of C3H7BO3, a gasoline additive. The products of combustion are C

nexus9112 [7]

Answer:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

Explanation:

For balancing a chemical equation, we should apply the law of conversation of mass. It states that the no. of atoms in the reactants side is equal to that of the products side.

So, the balanced equation:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

It is clear that 2.0 moles of C₃H₇BO₃ is completely burned in 8 m oles of oxygen and produce 6 moles of CO₂, 7 moles of H₂O and 1 mole of B₂O₃.

7 0

3 years ago

The gauge pressure inside a vessel is ‐40kPa, at an elevation of 5000m. a) What is the absolute pressure? b) At this elevation,

hoa [83]

Answer:

a) Pabs = 48960 KPa

b) T = 433.332 °C

Explanation:

Pabs = Pgauge + d*g* h

∴ d = 1000 Kg/m³

∴ g = 9.8 m/s²

∴ h = 5000 m

∴ P gauge = - 40 KPa * ( 1000 Pa / KPa ) = - 40000 Pa; Pa≡Kg/m*s²

⇒ Pabs = - 40000 Kg/ms² + ( 1000 Kg/m³ * 9.8 m/s² * 5000 m )

⇒ Pabs = 48960000 Pa = 48960 KPa

a) at that height and pressure, we find the temperature at which the water boils by means of an almost-exponential graph which has the following equation:

P(T) = 0.61094 exp ( 17.625*T / ( T + 243.04 ))......P (KPa) ∧ T (°C)....from literature

∴ P = 48960 KPa

⇒ ( 48960 KPa / 0.61094 ) = exp ( 17.625T / (T+ 243.04))

⇒ 80138.803 = exp ( 17.625T / ( T + 243.04))

⇒ Ln ( 80138.803) = 17.625T / ( T + 243.04))

⇒ 11.292 * ( T + 243.04 ) = 17.625T

⇒ 11.292T + 2744.289 = 17.625T

⇒ 2744.289 = 17.625T - 11.292T

⇒ 2744.289 = 6.333T

⇒ T = 433.332 °C

3 0

3 years ago