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VashaNatasha [74]
3 years ago
14

A 2.0% (w/v) solution of sodium hydrogen citrate, Na2C6H6O7, which also contains 2.5% (w/v) of dextrose, C6H12O6, is used as an

anticoagulant for blood which is to be used for transfusions. What is the molarity of the sodium hydrogen citrate in the solution
Chemistry
2 answers:
castortr0y [4]3 years ago
4 0

Answer:

0.085M

Explanation:

Molecular weight of sodium hydrogen citrate=236.09 g

236.09 g of Na2C6H6O7 in 1000ml= 1 M

2%= 2 g in 1 litre (1000ml)

20/236.09=0.0847 M=0.085 M

tamaranim1 [39]3 years ago
3 0

Answer:

0.0847M is molarity of sodium hydrogen citrate in the solution

Explanation:

The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.

<em>Moles Na2C6H6O7:</em>

<em>Molar Mass:</em>

2Na: 2*22.99g/mol: 45.98g/mol

6C: 6*12.01g/mol: 72.01g/mol

6H: 6*1.008g/mol: 6.048g/mol

7O: 7*16g/mol: 112g/mol

45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol

Moles of 2g:

2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>

<em />

<em>Liters solution:</em>

100mL * (1L / 1000mL) = <em>0.100L</em>

<em>Molarity:</em>

8.473x10⁻³ moles / 0.100L =

<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>
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A 10 gram sample of water is heated to 105 ℃ and is mixed with a 25 gram sample of water cooled to 25℃ . What is the final tempe
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Answer:

The final temperature of the water mixture is 47.85°C

Explanation :

Given,

For Warm Water

mass = 10grams

Temperature = 105°C

For Cold Water

mass = 25grams

Temperature = 25°C

When a sample of warm water is mixed with a sample of cool water,

The energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

<h3>Qlost = Qgain</h3>

However,

Q = (mass) (ΔT) (Cp)

Cp = Specific heat of water = 4.184 J/Kg°C

So,

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

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The warmer water goes down from to 105°C to x, so this means its Δt equals 105°C − x. The colder water goes up in temperature, so its Δt equals x − 25℃

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Solving for x, we get

x = 47.85°C

Therefore, The final temperature of the water mixture is 47.85°C.

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