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Pertanyaan

Sebanyak 60 gram urea ( M r ​ = 60 ) ; 11,7 gram NaCl ( M r ​ = 58 , 5 ) ;dan 11,1 gram CaCl 2 ​ ( M r ​ = 111 ) dilarutkan ke dalam 1 liter air. Tentukan titik beku larutan tersebut. ( K f ​ air = 1 , 86 ∘ C m − 1 )

Sebanyak 60 gram urea ; 11,7 gram ; dan 11,1 gram  dilarutkan ke dalam 1 liter air. Tentukan titik beku larutan tersebut.
 

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Jawaban

titik beku larutannya .

titik beku larutannya begin mathsize 14px style negative sign 3 comma 162 space degree C end style.

Pembahasan

Pembahasan
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Faktor Van't Hoff masing-masing larutan: Molalitas campurannya: Titik beku campuran dari larutan tersebut, Jadi, titik beku larutannya .

Faktor Van't Hoff masing-masing larutan:
i space equals space left parenthesis n space plus space 1 right parenthesis alpha space plus space 1

i subscript urea equals 1 space non space elektrolit  Na Cl space rightwards arrow space Na to the power of plus sign space plus space Cl to the power of minus sign i space equals space left parenthesis 2 minus sign 1 right parenthesis cross times 1 space plus space 1 i space equals space 2  Ca Cl subscript 2 space rightwards arrow space Ca to the power of 2 plus sign space plus space 2 Cl to the power of minus sign i space equals space left parenthesis 3 minus sign 1 right parenthesis cross times 1 space plus space 1 i space equals space 3 


Molalitas campurannya:


begin mathsize 14px style m equals left parenthesis mol subscript urea middle dot i subscript urea and mol subscript NaCl middle dot i subscript NaCl and mol subscript Ca Cl subscript 2 end subscript middle dot i subscript Ca Cl subscript 2 end subscript right parenthesis cross times 1000 over p m equals left parenthesis begin inline style 60 over 60 end style cross times 1 plus begin inline style fraction numerator 11 comma 7 over denominator 58 comma 5 end fraction end style begin inline style cross times end style begin inline style 2 end style begin inline style plus end style begin inline style fraction numerator 11 comma 1 over denominator 111 end fraction end style begin inline style cross times end style begin inline style 3 end style begin inline style right parenthesis end style begin inline style cross times end style begin inline style 1000 over 1000 end style m begin inline style equals end style begin inline style left parenthesis end style begin inline style 1 end style begin inline style plus end style begin inline style 0 end style begin inline style comma end style begin inline style 2 end style begin inline style cross times end style begin inline style 2 end style begin inline style plus end style begin inline style 0 end style begin inline style comma end style begin inline style 1 end style begin inline style cross times end style begin inline style 3 end style begin inline style right parenthesis end style begin inline style cross times end style begin inline style 1 end style begin inline style m end style begin inline style equals end style begin inline style 1 end style begin inline style comma end style begin inline style 7 end style begin inline style space end style begin inline style molal end style end style 


Titik beku campuran dari larutan tersebut,


begin mathsize 14px style space space space space space space increment T subscript f double bond m middle dot K subscript f T subscript f degree minus sign T subscript f double bond m middle dot K subscript f italic space italic space italic space 0 minus sign T subscript f equals 1 comma 7 cross times 1 comma 86 space space space space space space space space space T subscript f equals minus sign 3 comma 162 space degree C end style 


Jadi, titik beku larutannya begin mathsize 14px style negative sign 3 comma 162 space degree C end style.

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