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Dimas [21]
3 years ago
9

In a certain medical test designed to measure carbohydrate tolerance, an adult drinks 6 ounces of a 40% glucose solution. When t

he test is administered to a child, the glucose concentration must be decreased to 30%. How much 40% glucose solution and how much water should be used to prepare 6 ounces of 30% glucose solution?
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
8 0

Answer:

4.5 ounces of 40% glucose solution and 1.5 ounces of water.

Explanation:

The new solution will be prepared by adding more solvent, so the relation between concentration and volume is given by:

C1xV1 = C2xV2

Where C is the concentration, V is the volume, 1 represents the initial solution and 2 the final solution.

We want 6 ounces of a solution which 0.3 glucose, so :

0.4xV1 = 0.3x6

0.4xV1 = 1.8

V1 = 4.5 ounces

So, it's necessary 4.5 ounces of the 40% glucose solution. The volume of water (Vw) will be the total volume less the volume of the solution:

Vw = 6 - 4.5

Vw = 1.5 ounces

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\bold{\huge{\orange{\underline{ Solution}}}}

\bold{\underline{ Given :- }}

  • <u>We </u><u>have </u><u>250g </u><u>of </u><u>liquid </u><u>water </u><u>and </u><u>it </u><u>needs </u><u>to </u><u>be </u><u>cool </u><u>at </u><u>temperature </u><u>from </u><u>1</u><u>0</u><u>0</u><u>°</u><u> </u><u>C </u><u>to </u><u>0</u><u>°</u><u> </u><u>C</u>
  • <u>Specific </u><u>heat </u><u>of </u><u>water </u><u>is </u><u>4</u><u>.</u><u>1</u><u>8</u><u>0</u><u>J</u><u>/</u><u>g</u><u>°</u><u>C</u>

\bold{\underline{ To \: Find :- }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the</u><u> </u><u>total</u><u> </u><u>number </u><u>of </u><u>joules </u><u>released</u><u>. </u>

\bold{\underline{ Let's \:Begin:- }}

<u>We </u><u>know </u><u>that</u><u>, </u>

Amount of heat energy = mass * specific heat * change in temperature

<u>That </u><u>is, </u>

\sf{\red{ Q = mcΔT }}

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>

\sf{ Q = 250 × 4.180 ×(0 - 100 )}

\sf{ Q = 250 × 4.180 × - 100 }

\sf{ Q = 250 × - 418}

\sf{\pink{ Q = - 104,500 J }}

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

\bold{\underline{ Now :- }}

<u>We </u><u>have </u><u>to </u><u>tell </u><u>whether </u><u>the </u><u>above </u><u>process </u><u>is </u><u>endothermic </u><u>or </u><u>exothermic </u><u>:</u><u>-</u>

Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic

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