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3 years ago

How long will it take a piece of strontium-90 (produced in a nuclear test blast) weighing exactly 1000.0 g to be

Chemistry

1 answer:

Nonamiya [84]3 years ago

7 0

Answer:

Time taken 172.8 years

Explanation:

Given data:

Total amount of strontium-90 = 1000.0 g

Amount left = 15.625 g

Half life of strontium = 28.8 years

Time taken = ?

Solution:

Number of half lives:

At time zero = 1000.0 g

At first half life = 1000.0g/2 = 500 g

At second half life = 500 g/2 = 250 g

At third half life = 250 g/ 2 = 125 g

At 4th half life = 125 g/ 2= 62.5 g

At 5th half life = 62.5 g/2 = 31.25 g

At 6th half life = 31.25 g / 2 = 15.625 g

Time taken:

Time taken = number of half lives × half life

Time taken = 6×28.8 years

Time taken = 172.8 years

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Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?

german

Answer : The correct option is, (A) $AlCl_3$

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Al^{3+}+3Cl^{-}$ = 1 + 3 = 4

(B) $KI$

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of $KI$ will be,

$KI\rightarrow K^{+}+I^{-}$

So, Van't Hoff factor = Number of solute particles = $K^{+}+I^{-}$ = 1 + 1 = 2

(C) $CaCl_2$

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^{-}$ = 1 + 2 = 3

(D) $MgSO_4$

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of $MgSO_4$ will be,

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+SO_4^{2-}$ = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, $AlCl_3$

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